If `x in [-1, 0)` then find the value of `cos^(-1) (2x^(2) -1) -2 sin^(-1) x`

Explanation

To solve the problem, we need to find the value of \( \cos^{-1}(2x^2 - 1) - 2\sin^{-1}(x) \) given that \( x \in [-1, 0) \).

### Step-by-step Solution:

1. **Substitute \( x \) with \( \cos(\theta) \)**:
Since \( x \) lies in the interval \([-1, 0)\), we can let \( x = \cos(\theta) \) where \( \theta \) is in the range \([\pi/2, \pi)\).

**Hint**: This substitution helps us relate \( x \) to angles in trigonometric functions.

2. **Express \( 2x^2 - 1 \)**:
Substitute \( x = \cos(\theta) \) into \( 2x^2 - 1 \):
\[ 2x^2 - 1 = 2(\cos^2(\theta)) - 1 = \cos(2\theta) \]

**Hint**: Use the double angle formula for cosine, \( \cos(2\theta) = 2\cos^2(\theta) - 1 \).

3. **Rewrite the expression**:
Now we can rewrite the original expression:
\[ \cos^{-1}(2x^2 - 1) - 2\sin^{-1}(x) = \cos^{-1}(\cos(2\theta)) - 2\sin^{-1}(\cos(\theta)) \]

**Hint**: Recognize that \( \cos^{-1}(\cos(2\theta)) \) will yield \( 2\theta \) or \( 2\pi - 2\theta \) depending on the range of \( \theta \).

4. **Determine the value of \( \cos^{-1}(\cos(2\theta)) \)**:
Since \( \theta \) is in \([\pi/2, \pi)\), \( 2\theta \) will be in \([\pi, 2\pi)\). Thus:
\[ \cos^{-1}(\cos(2\theta)) = 2\pi - 2\theta \]

**Hint**: The range of \( \cos^{-1} \) is \([0, \pi]\), so we need to adjust for the angle being in the second cycle.

5. **Calculate \( \sin^{-1}(\cos(\theta)) \)**:
We know that:
\[ \sin^{-1}(\cos(\theta)) = \frac{\pi}{2} - \theta \]
Therefore:
\[ 2\sin^{-1}(\cos(\theta)) = 2\left(\frac{\pi}{2} - \theta\right) = \pi - 2\theta \]

**Hint**: Use the co-function identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \).

6. **Combine the results**:
Now substituting back into our expression:
\[ \cos^{-1}(2x^2 - 1) - 2\sin^{-1}(x) = (2\pi - 2\theta) - (\pi - 2\theta) \]
Simplifying this gives:
\[ = 2\pi - 2\theta - \pi + 2\theta = \pi \]

**Hint**: Carefully combine like terms to simplify the expression.

### Final Answer:
Thus, the value of \( \cos^{-1}(2x^2 - 1) - 2\sin^{-1}(x) \) is \( \pi \).